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Question
f(x) = x3 – 2x2 – x + 3 in [0, 1]
Solution
We have, f(x) = x3 – 2x2 – x + 3 in [0, 1]
Since, f(x) is a polynomial function it is continuous in [0, 1] and differentiable in (0, 1)
Thus, conditions of mean value theorem are satisfied.
Hence, there exists a real number c ∈ (0, 1) such that
f'(c) = `("f"(1) - "f"(0))/(1 - 0)`
⇒ 3c2 – 4c – 1 = `([1 - 2 - 1 + 3] - [0 + 3])/(1 - 0)`
⇒ 3c2 – 4c – 1 = –2
⇒ 3c2 – 4c + 1 = 0
⇒ (3c – 1)(c – 1) = 0
⇒ c = `1/3 ∈ (0, 1)`
Hence, the mean value theorem has been verified.
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